In front of a demon called Vishmaasur, 1000 people are lined up. The odd number in that line [e.g. 1,3,5,7,9] Eats poisonous people all around. After that, the queue of the remaining men remains. Now he eats poisonous people in the queue. In doing so, the queue gets shorter and eventually a person is left. So how much does that remaining man stand in the queue of the original 1000 people? Look at the number of people standing by trying.
Answers
Answer:
If all the odd peoples are removed and again the odd numbers in the remaing ones are removed the ans will be 512
512th numbered Person left at last
Step-by-step explanation:
Odd numbers being eaten ( 1, 3 , 5 , 7 )
1 , 2 ,3 , 4 , 5 , 6 ........................................................., 996 , 997 , 998 , 999 , 1000
After Eating 1st remaining
2 , 4 , 6 ................................................................, 996 , 998 , 1000
= 2( 1 , 2 , 3 ....................................................., 498 , 49 9 , 500)
after Eating 2nd remaining
= 2 ( 2 , 4 , ....................................................., 498 , 500)
= 2 * 2 ( 1 , 2 , ................................................, 249 , 250 )
= 4 ( 1 , 2 , ................................................, 249 , 250 )
after Eating 3rd remaining
= 4 ( 2 , 4 , ....................................................., 250 )
= 8 ( 1 , 2 .................................................... , 125)
after next
= 16 ( 1 , 2 , ........................................., 62)
after next
= 32 ( 1 , ............................................. 31 )
after next
= 64 ( 1 , ................................. . 15)
after next
= 128 ( 1 , ............................... , 7)
after next
= 256 ( 1 , 2 , 3 )
after next
= 256 ( 2)
= 512
512th is the assigned number of the person which is left at last
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